Problem: We are given that $\dfrac{dy}{dx}=x^2-2y$. Find an expression for $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$. $\dfrac{d^2y}{dx^2}=$
Notice that the equation defines $y$ implicitly—we don't have an explicit expression for $y$ in terms of $x$. So we will have to use implicit differentiation. If we differentiate the equation once, we will have $\dfrac{d^2y}{dx^2}$ on the left-hand side of the equation. This is what we get after we differentiate each side of the equation: $\dfrac{d^2y}{dx^2}=2x-2\dfrac{dy}{dx}$ The expression we got for $\dfrac{d^2y}{dx^2}$ isn't in terms of only $x$ and $y$ because it includes $\dfrac{dy}{dx}$ in it. Let's plug the original equation ${\dfrac{dy}{dx}=x^2-2y}$ in the new equation to obtain an expression in terms of only $x$ and $y$ : $\begin{aligned} \dfrac{d^2y}{dx^2}&=2x-2{\dfrac{dy}{dx}} \\\\ &=2x-2\left({x^2-2y}\right) \\\\ &=2x-2x^2+4y \end{aligned}$ In conclusion, this is an expression for $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$ : $\dfrac{d^2y}{dx^2}=2x-2x^2+4y$